My Approach

We always use nested loops for printing the patterns.

For the outer loop, we count the number of lines/rows and loop for them.
Next, for the inner loop, we focus on the number of columns and somehow connect them to the rows by forming a logic such that for each row we get the required number of columns to be printed.
We print the character inside the inner loop.
Observe symmetry in the pattern or check if a pattern is a combination of two or more similar patterns.

To analyze the time complexity of this program, let's break it down step by step:

Outer Loop (i): The outer loop runs N times, where N is the input parameter.
Inner Loop (j): This loop runs from the ASCII value of 'A' to the ASCII value of 'A' + N - i. In the worst case, when i is at its minimum value of 1, the inner loop will iterate from 'A' to 'A' + N - 1.

Let's calculate the total number of iterations of the inner loop across all iterations of the outer loop:

In the first iteration of the outer loop, the inner loop runs for N iterations.
In the second iteration of the outer loop, the inner loop runs for N - 1 iterations. And so on, until the last iteration of the outer loop where the inner loop runs for 1 iteration.

The sum of iterations is:

N + (N - 1) + (N - 2) + ... + 1

This sum is known as the sum of the first N natural numbers, which can be represented as $N*(N+1)/2$.

Therefore, the overall time complexity of the program is $O(N*(N+1)/2)$, which simplifies to $O(N^2)$ in big O notation.

where N is the number of rows/lines (horizontally).

The additional space used by the code is mainly for temporary variables and the input values.
The memory used by the nested loops is constant and does not depend on the input size N.
So, the space complexity of the code is $O(1)$, constant space complexity.

Provide feedback